Processing gain for wideband noise in DSSS Systems
As described by the OP; with Direct Sequence Spread Spectrum, narrow band data is spread over a wider bandwidth, and then despread back to a narrow bandwidth in the receiver. In the despreading process, a narrow interferor will be spread allowing for a processing gain relative to the demodulated data (the narrow band interferor gets spread over a wider bandwidth while the wideband signal gets despread to a narrow bandwidth). The processing gain applies directly to wideband noise as well. Wideband noise (white noise with consideration to the bandwidth of interest) results in each received sample being independent of each other. The despreading is done by multiplying the received waveform with a spreading code and then accumulating the result. This multiplication does not change the fact that each noise sample will still be independent of the next, but it does change the signal samples that were previously scrambled to all be in alignment: for example with DSSS using BPSK, the data is spread to be in pseudo random sequence of +1 and -1 symbols. After despreading, each symbol becomes all +1 (or all -1 depending on data bit), when these sum over $N$ samples, the signal level grows by $N$, or in dB specifically $20\log_{10}(N)$, while the independent noise samples on the other hand (as a noise component on each of these samples) will grow in standard deviation as a root-sum-square: $\sqrt{x_1^2+x_2^2 +x_3^2 + \ldots}$ or in dB as $10\log_{10}(N)$. The the processing gain as the difference in SNR will be $10\log_{10}(N)$.
General noise floor elevation in any receiver
If you elevate the noise density in the bandwidth of any modulated waveform with independent uncorrelated white (over the bandwidth of interest) noise so that the noise after all processing is elevated, then the SNR will degrade accordingly by that elevation, dB for dB. The noise floor in question however is not thermal noise at the input to the receiver, but it would be that plus the actual noise figure of the receiver (which includes all processing, not just the noise of the LNA alone); the input referred noise for the signal deep in the receiver just prior to decision.
Further, I believe the derived SNR at sensitivity is less than -10 dB, and to know what this is, the noise figure of the receiver must be known.
Explaining this: thermal noise over 500 MHz of bandwidth is given at room temperature by $kTB$ ($k$ as Bolzmann's constant, $T$ as Temperature in Kelvin and $B$ as Bandwidth in Hz) in $dBm$ (dB relative to one mW) quantities as $-174 \text{ dBm / Hz } + 10 Log10(500E6)= -87$ dBm total noise in 500 MHz of BW. The sensitivity is given as -98 dBm implying a minimum SNR to meet the performance requirements (such as bit error rate or some other related quality metric). If the receiver Noise Figure (including ALL noise in the receiver through final demodulation, not just the LNA), was impossibly 0 dB, then this received SNR level would be -98-(-87) = -11 dB. Since we know the Noise Figure must be some positive number (reasonably 2 to 6 dB, or possibly higher depending on quality of receiver), the actual noise level referred back to the receiver input as it would be relative to the signal level at decision will be elevated by the Noise Figure above -87 dBm. For example if the noise figure was 5 dB, the input referred noise would be -82 dBm, and then given a sensitivity of -98 dBm, the SNR at sensitivity would be -16 dB.
If additional noise was added at the input, if independent it would sum in power with the receiver noise floor, as well as the receiver noise contributions that resulted in the noise figure, to degrade the SNR accordingly: If the noise figure was 3 dB, this would mean the net contributions of all receiver noise exactly match in power the thermal noise floor, thus raising the input referred noise floor by 3 dB. If we for example added an additional noise spread evenly across the 500 MHz receiver bandwidth that was at -84 dBm, which itself is 3 dB above the thermal noise floor, then this would double the total noise further, elevating it another 3 dB when combined with the thermal noise and noise of the receiver. If we added noise that was at -94 dBm, or 7 dB below the thermal noise floor and 10 dB below the combined thermal and receiver noise, then the resulting total noise would increase by another 0.4 dB as given by the power sum converted back to dB:
$$10Log_{10}(1+10^{-10/10})= 0.4 \text { dB}$$
Where the 1 is $10^{0/10}$ as the "0 dB" reference converted to power of the thermal noise plus receiver noise, and the -10 in $10^{-10/10}$ is an additional noise 10 dB lower converted to power. We sum the two and then convert it back to dB to see that the result is increased by 0.4 dB above the 0 dB reference level.
